Can you speak English for a lowly S/W Dev?

[treat2]

Can you speak English for a lowly S/W Developer, please... I have always had an interest in physics, and have done some reading, on it, and will do much more. I'm not a mathematician, nor a Physicist, but I can catch on to ANYTHING well explained!

Here is the preliminary to the question I have concerning the increase in mass, as velocity increases. BUT first, I want to point to some Web Pages that confirm that a photon DOES have mass, as I'm not sure if it is part of my confusion or not.

1) "What is the Mass of a Photon?" See:
http://math.ucr.edu/home/baez/physi...hoton_mass.html

2) For a discussion of it, in another String Theory Forum, See:
http://superstringtheory.com/forum/...sages10/23.html
(Many other Web Sites can be found stating the Mass of a Photon).

If I recall correcly, I read on one Web Site that the mass of a photon is estimated at 10^-48th of the mass of an electron.

Now for my questions... THIS MUST BE VERY BASIC STUFF FOR YOU FOLKS
PLEASE excuse my ignorance.

1) What is meant by "infinite mass", when speaking of the mass of an object traveling at light speed.

2) Since a photon is traveling at light speed, why does it not have infinite mass; if not infinite mass, then at least the mass of a planet?

3) Along with infinite mass should come an infinite gravitational pull. So, why wouldn't an object with "infinite mass", have "infinite
gravity", which leads to two questions:
3a) What is infinite gravity?
3b) Why wouldn't the gravitational pull of the photon (which is supposed to have "infinite mass"), cause the entire planet to be swallowed into it, just like a Black Hole would do?

Again, I appologize for my ignorance, but these are the words used in physics books I've read, and I'm just following what I think is supposed to happen for a particle with mass traveling at light speed.
Thanks for your patience. I look forward to reading a reply that doesn't flame my ignorance. Thanks again.

 

[selfAdjoint]

I'll just do this bit.

If I recall correcly, I read on one Web Site that the mass of a photon is estimated at 10^-48th of the mass of an electron.

That is their upper limit considering unpreventable experimental error, i.e. if the photon had any mass it couldn't be bigger than that. There are very strong theoretical reasons to suppose the photon has no mass:

1) If the photon had mass then electromagnetic radiation would have a longitudunal component, and to the limits of experimental error, it doesn't (this argument goes back to Isaac Newton, no less, and it is actually the basis for the measurement you mentioned).

2) The photon interactions in the relativistic quantum electrodynamics would come out wrong, and they wouldn't be able to calculate things like the Lamb shift and the anomalous magnetic moment of the electron as accurately as they do. This accuracy is the best that any physics theory has ever generated; calculation matches experiment to six decimal places.

So physicsits feel pretty sure the photon is massless, and experiment doesn't contradict them whatever some crank websites may assert. Every time the experimenters are able to increase their accuracy, that upper limit goes down, and experimentalists will tell you, there is NO EXPERIMENTAL EVIDENCE FOR PHOTON MASS.

 

[pmb_phy]

Originally posted by treat2
Can you speak English for a lowly S/W Developer, please... I have always had an interest in physics, and have done some reading, on it, and will do much more. I'm not a mathematician, nor a Physicist, but I can catch on to ANYTHING well explained!

Here is the preliminary to the question I have concerning the increase in mass, as velocity increases. BUT first, I want to point to some Web Pages that confirm that a photon DOES have mass, as I'm not sure if it is part of my confusion or not.

You're confusing the two different uses of the term "mass." That link you provided explains this difference. An increase in speed yielding an increase in mass refers to what is sometimes known as "relativistic mass." This mass can be thought of as the ratio of the magnitude of a particle's momentum to its speed, i.e. m = p/v. When it is said that light has zero mass it refers to "proper mass" aka "rest mass."

Point by point

1) What is meant by "infinite mass", when speaking of the mass of an object traveling at light speed.

It means that, given a particle with a finite proper mass, m₀, as the particle's speed approaches the speed of light its momentum increases without bound. This follows from the relations

$$\mathbf{p} = m\mathbf{v}$$

$$m = \gamma m_{0} = \frac {m_{0}} { \sqrt{1-v^{2}/c^{2}} }$$

2) Since a photon is traveling at light speed, why does it not have infinite mass; if not infinite mass, then at least the mass of a planet?

Since both the momentum and the velocity of light are finite then so is the ratio p/c = finite.

3) Along with infinite mass should come an infinite gravitational pull. So, why wouldn't an object with "infinite mass", have "infinite
gravity",..

That is true. If you're in a gravitational field in which you measure a finite gravitational acceleration of a falling particle then if you change to a frame of reference which is moving with respect to that frame then the magnitude of the gravitational acceleration as measured in this "moving" frame will increase. For example: Suppose you have a large sheet of matter which generates a uniform gravitational field. Assuming the pressure of the matter is small compared to the mass density then the gravitational field in the rest frame of the matter will give rise to a gravitational acceleration which is proportional to the surface mass density (mass per unit area) of the sheet. Let this acceleration be g₀. Now transform to a frame of reference moving with respect to the sheet. There will be an increase in the relativistic mass of each portion of the sheet and a decrease in the volume of each portion. There will therefore be a gamma factor for each. In the moving frame the instaneous gravitational acceleration (asusming falling particle has zero speed) will then be g where

$$g = \gamma^{2} g_{0} = \frac {g_{0}} {1-v^{2}/c^{2} }$$

As the speed of the sheet goes to c then g goes to infinity.

That should take care of the rest of the questions.

I hope that helps.

 

[HallsofIvy]

Originally posted by treat2[/b]
1) What is meant by "infinite mass", when speaking of the mass of an object traveling at light speed.

Well, I would interpret it as saying that any object with mass can't travel at light speed!

2) Since a photon is traveling at light speed, why does it not have infinite mass; if not infinite mass, then at least the mass of a planet?

Yes, any object with non-zero "rest-mass" would have infinite mass when traveling at light speed- which is why it can't! A photon, however, has no rest-mass and so has no mass at all even though it is moving at light speed.

 

[GRQC]

Originally posted by HallsofIvy
Yes, any object with non-zero "rest-mass" would have infinite mass when traveling at light speed- which is why it can't! A photon, however, has no rest-mass and so has no mass at all even though it is moving at light speed. [/B]

No. This is incorrect. Mass is a relativistic invariant quantity. The biggest confusion I have seen on this board in the past 2 days is the distinction between mass (magnitude of a 4-dimensional vector) and energy (time component of that vector).

The energy required to accelerate the object will grow to infinity as the velocity $\rightarrow c$, but the mass stays the same. That is, energy and momentum are relative quantities, but mass is not.

A very good exposition on this confusion can be found in:

Taylor and Wheeler, Spacetime Physics: Introduction to Special Relativity (2nd ed).

See pp. 246-251, "Dialog: Use and Abuse of the Concept of Mass"

All ye who covet the phrase "relativistic mass" should read and heed...

 

[pmb_phy]

Originally posted by HallsofIvy
Well, I would interpret it as saying that any object with mass can't travel at light speed!


Yes, any object with non-zero "rest-mass" would have infinite mass when traveling at light speed- which is why it can't! A photon, however, has no rest-mass and so has no mass at all even though it is moving at light speed.

To be precise one would say that if a particle has a non-zero proper mass then it can't be accelerated to the speed of light from a speed which is greater than or less than the speed of light. However if a partilce is "born" traveling at a speed greater than or less than the speed of light then it will always travel at a speed greater than or less than the speed of light respectively.

Particles are classified according to this scenario

Tardyon - Speed always less than c
Luxon - Speed always equal to c
Tachyon - Speed always greater than c

where "c" refers to the speed of light as measured locally and the speed mentioned above refers to the locally measured speed.

 

[meddyn]

Treat2, I also develop software for the mundane world of invasive cardiology and thoracic surgery measuring and interpreting biological anomalies. Reading the forum over time has led to a firm understanding that questions agreeing with ordinary logic are often illogical ("relative" to an ordinary viewpoint) mathematically. It seems quite simple (relative to ordinary logic) that a galaxy traveling away from us at 100 miles per second is emitting light in the opposite direction at C + 100 miles per second and emitting light back at us at C - 100 miles per second.

The answer, of course, depends on where the viewpoint is and some really impressive formulas that say it is and is not true, depending. And to embelish your questions, if the photon has no mass then "ordinary logic" ask how can light be gravitationally "lensed" around a large mass body. Doesn't gravity act on mass? I think the reply there really does create some interest (and good formulas too).

One thing I was very impressed with recently was a "Law" I heard in a course on Emergency Medicine. "Do not rely on your logic in treatment; rely on your training and education. The injury/disease vs. treatment that is appropriate may not be logical at your level". So, at my level, I just continue to admire and be impressed with this forum while learning some neat stuff.

 

[pmb_phy]

Originally posted by meddyn
It seems quite simple (relative to ordinary logic) that a galaxy traveling away from us at 100 miles per second is emitting light in the opposite direction at C + 100 miles per second and emitting light back at us at C - 100 miles per second.

It might seem that way but that has never been observed. People made the assumption for a long time that the speed of light was a function of the velocity of the source emitting the light. However all experimental efforts to verify this assumption met with failure. For this and other reasons in 1905 Einstein postulated that the speed of light in vacuum is indpendant of the motion of the source. As such the speed of the source does not play a role in the speed of light emitted from the source. To be exact the postulate was phrased in his paper as

...light is always propagated in empty space with a definite velocity c which is independant of the state of motion of the emitting body.

Predictions based on that assumption have been borne out in all experiments executed to date.

And to embelish your questions, if the photon has no mass then "ordinary logic" ask how can light be gravitationally "lensed" around a large mass body. Doesn't gravity act on mass? I think the reply there really does create some interest (and good formulas too).

This confusion arises when once fails to distinguish the difference between mass, m, (relativistic mass) and proper mass. m₀.

The mass (i.e. relativisitc mass), m, of light is the ratio of the momentum of light to the speed of light. The proper mass of the photon is zero.

Think of proper mass as you would proper time and mass as you would time.

 

[franznietzsche]

Originally posted by GRQC
No. This is incorrect. Mass is a relativistic invariant quantity. The biggest confusion I have seen on this board in the past 2 days is the distinction between mass (magnitude of a 4-dimensional vector) and energy (time component of that vector).

The energy required to accelerate the object will grow to infinity as the velocity $\rightarrow c$, but the mass stays the same. That is, energy and momentum are relative quantities, but mass is not.

A very good exposition on this confusion can be found in:

Taylor and Wheeler, Spacetime Physics: Introduction to Special Relativity (2nd ed).

See pp. 246-251, "Dialog: Use and Abuse of the Concept of Mass"

All ye who covet the phrase "relativistic mass" should read and heed...

Get off your bloody pedastal, you know what we mean, we are referring to realtivistic mass. Bottom line: Kinectic energy increases the effective mass, the mass that one must deal with in equations such as $$F=ma$$ with $$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$. You insisting on saying that $$F=\gamma*m_0a$$ with [tex] \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} [\tex] is exactly equivalent. IT MAKES ABSOLUTELY NO DIFFERENCE. So just get over it.

 

[GRQC]

Originally posted by franznietzsche
Get off your bloody pedastal, you know what we mean, we are referring to realtivistic mass. ... snip ...

It makes a very big difference. There is no such thing as relativistic mass. If you're going to learn proper relativity theory, you must know the difference. There is a very important difference between mass and energy-momentum that must be acknowledged.

In fact, I would avoid $\gamma m_0 a$ for relativistic force, and use $\frac{dp}{dt}$ instead. This removes any confusion about "mass" from the picture, and introduces the more appropriate relativistic variable, 3-momentum.

You cannot write $m = \gamma m_0$ because rest mass is a relativistic invariant! Have you ever see relativity texts talk about "relativistic proper time", $\tau = \gamma \tau_0$?

I'm afraid I'm not to one who has to get over it.

 

[pmb_phy]

Originally posted by franznietzsche
Bottom line: Kinectic energy increases the effective mass, ...

Yes. That is 100% correct.

..the mass that one must deal with in equations such as $$F=ma$$ with $$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$.

What do you mean by this? The definition of force is not f = ma. It's f = dp/dt where p = mv. It's this later expression which defines mass, not the first.

You insisting on saying that $$F=\gamma*m_0a$$ with $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ is exactly equivalent.

That is a relationship between the transverse component of force, the transverse acceleration and the transverse mass = relativistic mass. In that expression the F is transverse component of force and so it does not hold when F is the total force.

 

[franznietzsche]

Originally posted by pmb_phy

What do you mean by this? The definition of force is not f = ma. It's f = dp/dt where p = mv. It's this later expression which defines mass, not the first.

$$p = mv m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} F = \frac{dp}{dt} = \frac{dm}{dt}v + m\frac{dv}{dt}$$

now assuming mass is invariant, this reduces to F = ma. For our case where mass increases only with velocity, and not independently of time we get:

$$F = \left(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\right)\frac{dv}{dt}$$

Which is relativistic mass times acceleration. Ergo, F = ma. That is what i meant. Using [tex] \frac{dp}{dt} [tex] is unnecessary except in the cases where mass varies dependent on time, and those cases are compartively rare, at least that i know of. Also it is far easier to experimentally measure rest mass, then apply the relevant Lorentz transformation than it is to measure the change in momentum. F=ma is easier to use for experimental purposes.

You cannot write $$m =\gammam_0$$ because rest mass is a relativistic invariant

The use of that equation implies that rest mass is invariant. However when dealing with forces acting on an object, rest mass is useless(except in the exception of stationary objects, but since a force is acting on them they will not long remain stationary, we must know the the relativistic mass.

Edit: Why is the tex not working? and the quote for that matter? Its been working on/and off all day for me...

 

[GRQC]

Originally posted by franznietzsche

However when dealing with forces acting on an object, rest mass is useless(except in the exception of stationary objects, but since a force is acting on them they will not long remain stationary, we must know the the relativistic mass.

No, you use the time-derivative of relativistic momentum to define force. The notion of mass is out of the picture. You're strictly dealing with energy and momentum when using Lorentz transformations. Your derivation is erroneous because you a priori assume that
$m = \gamma m_0$, which is false. You must use four-vector quantities when dealing with frame transformations. No two ways about it; that's the point of special relativity! Mass is the invariant magnitude of 4-momentum.

Again, have you ever seen someone Lorentz transform proper time?

Edit: Why is the tex not working? and the quote for that matter? Its been working on/and off all day for me... [/B]

Because you're using the wrong slash. You should be closing them with '/' instead of '\'.

 

[pmb_phy]

Originally posted by franznietzsche

..
Now assuming mass is invariant, this reduces to F = ma. For our case where mass increases only with velocity, and not independently of time we get:

First off please note that you're using the term "invariant" to mean "does not depend on time". The term "invariant" in relativity is usually taken to mean "does not change upon a change in coordinates."

The only way that dm/dt can be zero is for a = 0.

Which is relativistic mass times acceleration. Ergo, F = ma.

Do this calculation out explicitly and you'll see what I mean and note that v = v(t).

That is what i meant. Using $$\frac{dp}{dt}$$ is unnecessary except in the cases where mass varies dependent on time, and those cases are compartively rare, at least that i know of.

Note that if m is a function of v then for an accelerating particle v = v(t) and therefore m = m(t). So this is not a special case but is the general case. I.e. the mass of a particle changes when there is a force acting on it.

Also it is far easier to experimentally measure rest mass, then apply the relevant Lorentz transformation than it is to measure the change in momentum. F=ma is easier to use for experimental purposes.

Actually for subatomic particles its just as easy to measure relativisitc mass as it is to measure rest mass. Probably easier as a matter of fact. Take the case of an electron as an example. Shoot an electron into a uniform magnetic field perpendicular to the field lines and measure the radius of curvature of the path and the cyclotron frequency and with the value of the magnetic field strength you can calculate the value of the relativistic mass. See

http://www.geocities.com/physics_world/sr/cyclotron.htm

You cannot write $$m =\gammam_0$$ because rest mass is a relativistic invariant

That is an incorrect statement. The term relativistic mass is the name given to the quantity m = m₀/sqrt[1-(v/c)^2]. Proper mass is the invariant quantity in that relation, not relativistic mass (although there is a context in which one can say that relativistic mass is a scalar). In relativity the qualifier "relativistic" does not mean "invariant" it means "precise at all speeds less than the speed of light." That is the reason that

$$\mathbf{p} = \frac {m_{0}\mathbf{v}} { \sqrt{1-v^{2}/c^{2} } }$$

is referred to as relativistic momentum (e.g. Ohanian 2001).

Edit: Why is the tex not working? and the quote for that matter? Its been working on/and off all day for me...

You have the slash going the wrong way when you end the "tex". i.e. it should read "[/tex]" but you're writing it as "[\tex]"

 

[meddyn]

So, Treat2, after carefully reviewing and testing the replys I worked out a formula based on relativity:

U = (MC^2) - E

Let U = Understanding

 

[franznietzsche]

Originally posted by meddyn
So, Treat2, after carefully reviewing and testing the replys I worked out a formula based on relativity:

U = (MC^2) - E

Let U = Understanding

refreshing humor indeed.

 

[DW]

Originally posted by treat2
BUT first, I want to point to some Web Pages that confirm that a photon DOES have mass, as I'm not sure if it is part of my confusion or not.

1) "What is the Mass of a Photon?" See:
http://math.ucr.edu/home/baez/physi...hoton_mass.html

Actually the site says the opposite of what you claim. In specific
"Does light have mass?
The short answer is "no", but it is a qualified "no" because there are odd ways of interpreting the question which could justify the answer 'yes'."
http://tinyurl.com/ywvww

The only other of the "pages" you list as far as I can tell is
http://tinyurl.com/22r8d
which is not a verification, but a layman discussion board.

(Many other Web Sites can be found stating the Mass of a Photon).

And they are wrong if aren't listing a null result.

If I recall correcly, I read on one Web Site that the mass of a photon is estimated at 10^-48th of the mass of an electron.

No that is the maximum mass that it could have been due to limitations on the experiments precision.

1) What is meant by "infinite mass", when speaking of the mass of an object traveling at light speed.

Mass does not change with speed. Things that travel at the speed of light don't have infinite mass. They have zero mass.

Since a photon is traveling at light speed, why does it not have infinite mass...

Why would it? Things that travel at the speed of light have zero mass.

Along with infinite mass should come an infinite gravitational pull.

Nothing has infinite mass so it wouldn't matter, but even so relativistic gravitation is not given by Newton's law of gravitation.

What is infinite gravity?

Something that doesn't exist.

Why wouldn't the gravitational pull of the photon (which is supposed to have "infinite mass"),

Its not supposed to have any such thing. Its supposed to have zero mass.

cause the entire planet to be swallowed into it, just like a Black Hole would do?

Since it has zero mass and its gravitation isn't given by Newton's law of gravitation, why should that be so?

Again, I appologize for my ignorance, but these are the words used in physics books I've read, and I'm just following what I think is supposed to happen for a particle with mass traveling at light speed.

At the least you have provided an excellent example of why in the modern relativistic terminology the the concept of "relativistic mass" has been done away. I recommend getting better books. If it must be at the undergrad level then I recommend any recent text by Serway that has a chapter on at least special relativity.

 

[DW]

Originally posted by pmb_phy
You're confusing the two different uses of the term "mass." ... An increase in speed yielding an increase in mass refers to what is sometimes known as "relativistic mass."

This is a good example of why you should stop using this obsolete concept.

m = p/v

Correction, the mass m is not the relativistic mass M = p/v.

When it is said that light has zero mass it refers to "proper mass" aka "rest mass."

It means just exactly what it said, zero mass.

It means that, given a particle with a finite proper mass, m₀,

Should be written just m as the zero subscript is meaningless. The mass is invariant. It is the same for every frame, not just the value of the mass for the rest frame.

$$\mathbf{p} = m\mathbf{v}$$

Should be $$P^i = mU^i = \gamma mu^i$$.

$$m = \gamma m_{0} = \frac {m_{0}} { \sqrt{1-v^{2}/c^{2}} }$$

Should be $$E = \gamma mc^2 = \frac {mc^{2}} { \sqrt{1-v^{2}/c^{2}} }$$

If you're in a gravitational field in which you measure a finite gravitational acceleration of a falling particle then if you change to a frame of reference which is moving with respect to that frame then the magnitude of the gravitational acceleration as measured in this "moving" frame will increase.

And by that line of reasoning if you change to a frame of reference which is moving with respect to the second and just happen to do get back to the original motion state by that boost it must increase yet again! Every frame is in motion with respect to some frame, even your original frame.

(snipped the rest as its irrelevant to his question at hand)

 

[DW]

Originally posted by pmb_phy
This confusion arises when once fails to distinguish the difference between mass, m, (relativistic mass) and proper mass. m₀.

Like you are leading people to do right now. It should be
...between mass $$m$$, and (relativistic mass) $$M = \gamma m$$.
And you shouldn't even be using the latter.

The mass (i.e. relativisitc mass), m,

Correction, ...relativisitc mass), M...

The proper mass of the photon is zero.

Correction - The mass of the photon is zero.

Think of proper mass as you would proper time and mass as you would time.

Argument by analogy does not constitute a logical argument. As I told you befor, the relativity of spacetime coordinates is not analogous to the invariance of the physics.

 

[DW]

Originally posted by franznietzsche
Get off your bloody pedastal, you know what we mean, we are referring to realtivistic mass. Bottom line: Kinectic energy increases the effective mass, the mass that one must deal with in equations such as $$F=ma$$ with $$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$. You insisting on saying that $$F=\gamma*m_0a$$ with $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ is exactly equivalent. IT MAKES ABSOLUTELY NO DIFFERENCE. So just get over it.

This is why you need to stop referring to relativistic mass and using it at all. Your equation is just plain WRONG.

 

[DW]

Originally posted by pmb_phy
The definition of force is not f = ma. It's f = dp/dt where p = mv.

Correction - ordinary force. F = mA does work as the four vector law of motion.

It's this later expression which defines mass, not the first.

No it doesn't. It defines ordinary force as you were almost saying correct above. It can't define both sides of the equation without being circular.

That is a relationship between the transverse component of force, the transverse acceleration and the transverse mass = relativistic mass.

No, the equation
$$f = m_{long} a_{long} + m_{trans} a_{trans}$$ where you are in essence defining $$m_{long}$$ as $$\gamma ^{3}m$$ and $$m_{trans}$$ as $$\gamma m$$is also wrong.

 

[DW]

Originally posted by franznietzsche
$$p = mv$$

Should be
$$P^i = mU^i = \gamma mu^i$$

$$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

Should be $$E = \frac{mc^{2}}{\sqrt{1-\frac{v^2}{c^2}}}$$

$$F = \frac{dp}{dt} = \frac{dm}{dt}v + m\frac{dv}{dt}$$

Should be $$f^i = \frac{dP^i}{dt} = \frac{d\gamma }{dt}mu^i + \gamma m\frac{du^i}{dt}$$

now assuming mass is invariant, this reduces to F = ma.

No it doesn't. It reduces to $$f^i = \gamma m[a^i + \gamma ^{2}\frac{\vec{u} \cdot \vec{a}}{c^2}u^i]$$

For our case where mass increases only with velocity, and not independently of time we get:
$$F = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\frac{dv}{dt}$$

This equation is just plain wrong and conclusions based on it are invalid.

Using $$\frac{dp}{dt}$$ is unnecessary except in the cases where mass varies dependent on time, and those cases are compartively rare, at least that i know of.

Rockets are not rare.

Also it is far easier to experimentally measure rest mass, then apply the relevant Lorentz transformation than it is to measure the change in momentum.

The mass is the mass according to every frame. It is not any different for a rest frame than any other frame. It is INVARIANT.

The use of that equation implies that rest mass is invariant.

Mass is invariant.

However when dealing with forces acting on an object, rest mass is useless(except in the exception of stationary objects, but since a force is acting on them they will not long remain stationary, we must know the relativistic mass.

No, the mass is not just the mass for the rest frame as it is invariant. The "relativistic mass" is an obsolete concept.

 

[DW]

Originally posted by pmb_phy
The only way that dm/dt can be zero is for a = 0.

Whether dm/dt is zero has nothing to do with whether a = 0.

Note that if m is a function of v then for an accelerating particle v = v(t) and therefore m = m(t).

m is not a function of v. It is invariant.

So this is not a special case but is the general case. I.e. the mass of a particle changes when there is a force acting on it.

No, it is invariant.

Actually for subatomic particles its just as easy to measure relativisitc mass as it is to measure rest mass.

Should read, Actually for subatomic particles its just as easy to measure E as it is to measure the mass.

http://www.geocities.com/physics_world/sr/cyclotron.htm

Invalid, instead see-
http://www.geocities.com/zcphysicsms/chap3.htm

The term relativistic mass is the name given to the quantity m = m₀/sqrt[1-(v/c)^2].

Should read, The obsolete term relativistic mass was the name given to the quantity $$M$$ in $$M = \frac{m}{\sqrt{1 - (v/c)^2}}$$.

Proper mass is the invariant quantity in that relation, not relativistic mass

Should read, Mass is the invariant quantity in that relation, not the energy.

(although there is a context in which one can say that relativistic mass is a scalar)

Here you are making a reference to your statement already disproven at another board that energy was invariant as well as your argument also disproven that scalar implies and is implied by invariant.

$$\mathbf{p} = \frac {m_{0}\mathbf{v}} { \sqrt{1-v^{2}/c^{2} } }$$
is referred to as relativistic momentum

Should read, $$P^i = mU^i = \gamma mu^i = \frac{mu^i}{\sqrt{1 - \frac{u^2}{c^2}}}$$

 

[GRQC]

Thank you, DW. As a follow-up, I cite a few passages from "Spacetime Physics", by Wheeler and Taylor:

Does an isolated system have the same mass as observed in every inertial (free-float) reference frame?

Yes, given in terms of energy E and momentum p by $m^2=E^2-p^2$ in one frame, and $m^2 = E^\prime^2 - p^\prime^2$ in another frame. Mass of an isolated system is thus an
invariant.

...

Is the mass of a moving object greater than the mass of the same object at rest?

No. It is the same whether the object is at rest or in motion; the same for all reference frames.
...
The concept of "relativistic mass" is subject to misunderstanding. That's why we don't use it. First, it applies the name mass -- belonging to the magnitude of a 4-vector -- to ... the time component of a 4-vector. Second, it makes increase of energy of an object with velocity or momentum appear to be connected with some change in internal structure of the object. In reality, the increase in energy with velocity origniates not in the object but in the geometric properties of spacetime itself.
...
Mass has the same value in all frames, is invariant, no matter how the particle moves.

The takehome lesson is that mass is invariant. No such thing as relativistic mass, anymoreso than there is "relativistic spacetime interval".

 

[Nereid]

Originally posted by meddyn
Treat2, I also develop software for the mundane world of invasive cardiology and thoracic surgery measuring and interpreting biological anomalies. Reading the forum over time has led to a firm understanding that questions agreeing with ordinary logic are often illogical ("relative" to an ordinary viewpoint) mathematically. It seems quite simple (relative to ordinary logic) that a galaxy traveling away from us at 100 miles per second is emitting light in the opposite direction at C + 100 miles per second and emitting light back at us at C - 100 miles per second.

The answer, of course, depends on where the viewpoint is and some really impressive formulas that say it is and is not true, depending. And to embelish your questions, if the photon has no mass then "ordinary logic" ask how can light be gravitationally "lensed" around a large mass body. Doesn't gravity act on mass? I think the reply there really does create some interest (and good formulas too).

One thing I was very impressed with recently was a "Law" I heard in a course on Emergency Medicine. "Do not rely on your logic in treatment; rely on your training and education. The injury/disease vs. treatment that is appropriate may not be logical at your level". So, at my level, I just continue to admire and be impressed with this forum while learning some neat stuff.

LOL! Welcome Treat2, welcome meddyn!

Just for the record, in case this might get lost in all the discussion, however you do the math, no experiment yet has shown any inconsistency with SR or GR. Along with QED (crudely, the incorporation of SR into Quantum Mechanics), they are among the most stringently tested parts of physics. Yet, curiously, QM and GR cannot both be right. Personally, I find it simply awesome (and yes, some of the math is pretty cool, if not exactly intuitive).

 

[Omni]

I think you all missed the bit about "Can you speak English for a lowly S/W Developer"

 

[treat2]

Yeah. That was forgetten way back!

Originally posted by GRQC
Thank you, DW. As a follow-up, I cite a few passages from "Spacetime Physics", by Wheeler and Taylor:

The take home lesson is that mass is invariant. No such thing as relativistic mass, anymoreso than there is "relativistic spacetime interval".

OK. I'm referring back to your post at the end of Page 2 of this Thread. (I think.)

I went back again, and looked at the Physics FAQ I was referred to:

at: http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html

and cut out this small section, that begins with the question:
"Is there any experimental evidence that the photon has zero rest mass?"

Below is part of the answer from that page:

"If the rest mass of the photon was non-zero, the theory of quantum electrodynamics would be "in trouble" primarily through loss of gauge invariance, which would make it non-renormalizable; also,
charge-conservation would no longer be absolutely guaranteed, as it is if photons have vanishing rest-mass. However, whatever theory says, it is still necessary to check theory against experiment.
It is almost certainly impossible to do any experiment which would establish that the photon rest mass is exactly zero. The best we can hope to do is place limits on it. A non-zero rest mass would lead to a change in the inverse square Coulomb law of electrostatic forces. There would be a small damping factor making it weaker over very large distances. The behavior of static magnetic fields is likewise modified. A limit on the photon mass can be obtained through satellite measurements of planetary magnetic fields. The Charge Composition Explorer spacecraft was used to derive a limit of 6x10-16 eV with high certainty. This was slightly improved in 1998 by Roderic Lakes in a laborartory experiment which looked for anomalous forces on a Cavendish balance. The new limit is 7x10-17 eV. Studies of galactic magnetic fields suggest a much better limit of less than 3x10-27 eV but there is some doubt about the validity of this method."
---end quote---------

It states that photons have a "rest mass". It says the "...limit of" that was measured "with high certainty." AND "This was slightly improved in 1998" to be 7x10-17 eV.

Just talk English to me.

(The equations in this thread are not impressing me, and no one seems to be able to provide the equations for what I would expect should be easy for any Physics Major. LOL!)

When the FAQ says that photons have a mass of so-and-so, I'm not sure where they are contradicting me original question, OR EVEN THEMSELVES.

IS "eV" an "electon Volt"?

Did they SIMPLY DO A SUBTITUTION for a MEASUREMENT OF MASS (ABOVE),
and express MASS in terms of a MEASURMENT for ENERGY?

Because >>>>THAT<<<<< IS WHAT I'm guessing is the ENTIRE source of confusion (as far as I'm concerned).

Would someone get back to English, or do you want me to start talking in C# ICL, so ain't NOBODY going to get nuffin outta this thread! LOL!

The other aspect of confusion was cleared up, by saying that mass is invariant, and that the variant is Energy (which I happen to recall),
but not from this thread.) I won't even bother asking what Eo means.
(I'll just assume it is when nothing is moving.)

It's just my brain rotting after reading 3 pages of stuff with equations and terminology that no one agrees on, and precious little ENHLISH!

The "take home lesson" is in what I UNDERSTAND to be the CORRECT terminology to be using. Except, lots of ppl and articles don't bother doing it, and it just makes from more questions, than answers
(on my part). However, I accept that as being the correct terminology,
despite my not understanding all of the implications of it. Except that if mass doesn't change, there should not be an increase in gravity, as I asked about. Don't bother explaining it with equations.
Before you talk equations, there needs to be quite a bit more definitons.) Just a simple few sentences not using abbreviations, as you almost did, would explain the answers in a way that EVERYONE will understand. BTW. I assume a 4 vector are the 3 Dimensions, Plus Time as the 4th Dimension. GEEEZZ. I REALLY oue some of you folks some C# ICL, and let you ponder that stuff! LOL!

 

[GRQC]

Originally posted by treat2

It states that photons have a "rest mass". It says the "...limit of" that was measured "with high certainty." AND "This was slightly improved in 1998" to be 7x10-17 eV.

Careful, it does not say that photons have rest mass. It says that if they have (rest) mass, then it is constrained to be very, very, very small. As a contrast, the electron has a mass of 0.5 MeV (i.e. million eV).

By the way, yes, eV is an electron volt (unit of energy). In particle physics, mass is expressed this way because the numbers are more manageable then expressing mass in kg. When you see eV as mass, it actually means eV/c².

Back to the sermon: we used to think neutrinos were massless, but there is evidence to support at least some of them have mass (on the order of a few eV). Theory originally stated that they were massless, but this was relatively easy to modify to include mass (also, there were other observed phenomena -- the solar neutrino problem -- which could be explained if neutrinos had mass (something called neutrino oscillation).

The bottom line is, as you've ascertained from that website: theory would be in trouble if the photon had mass. And, this is a well-established theory (quantum electrodynamics), which has stood up to incredible experimental verification, so the chance that something like this could kill it is pretty remote. Unlike with neutrino physics, there is nothing we observe that could be explained by a massive photon. It's more or less a pipe dream of researchers who need papers to write (and want to stir up some trouble).

Again, reporting an upper bound on the mass doesn't mean it has mass. It is just a consequence of experimental verification. We can't measure exactly 0 mass, but the closer we get to 0 the more likely it is 0.

Just talk English to me.

Hope that was good enough.

 

[Nereid]

... and 7x10^-17 eV is how many kg?

Another way to express the upper limit of the mass of a photon: it's <2x10^-22 the mass of an electron.

 

[Janus]

Originally posted by Nereid
... and 7x10^-17 eV is how many kg?

.

1.3 x 10^{-52 kg}

 

[treat2]

Originally posted by GRQC
Careful, it does not say that photons have rest mass. It says that if they have (rest) mass, then it is constrained to be very, very, very small. As a contrast, the electron has a mass of 0.5 MeV (i.e. million eV).

By the way, yes, eV is an electron volt (unit of energy). In particle physics, mass is expressed this way because the numbers are more manageable then expressing mass in kg. When you see eV as mass, it actually means eV/c².

Back to the sermon: we used to think neutrinos were massless, but there is evidence to support at least some of them have mass (on the order of a few eV). Theory originally stated that they were massless, but this was relatively easy to modify to include mass (also, there were other observed phenomena -- the solar neutrino problem -- which could be explained if neutrinos had mass (something called neutrino oscillation).

The bottom line is, as you've ascertained from that website: theory would be in trouble if the photon had mass. And, this is a well-established theory (quantum electrodynamics), which has stood up to incredible experimental verification, so the chance that something like this could kill it is pretty remote. Unlike with neutrino physics, there is nothing we observe that could be explained by a massive photon. It's more or less a pipe dream of researchers who need papers to write (and want to stir up some trouble).

Again, reporting an upper bound on the mass doesn't mean it has mass. It is just a consequence of experimental verification. We can't measure exactly 0 mass, but the closer we get to 0 the more likely it is 0.



Hope that was good enough.

Originally posted by GRQC
Careful, it does not say that photons have rest mass. It says that if they have (rest) mass, then it is constrained to be very, very, very small. As a contrast, the electron has a mass of 0.5 MeV (i.e. million eV).

By the way, yes, eV is an electron volt (unit of energy). In particle physics, mass is expressed this way because the numbers are more manageable then expressing mass in kg. When you see eV as mass, it actually means eV/c².

Back to the sermon: we used to think neutrinos were massless, but there is evidence to support at least some of them have mass (on the order of a few eV). Theory originally stated that they were massless, but this was relatively easy to modify to include mass (also, there were other observed phenomena -- the solar neutrino problem -- which could be explained if neutrinos had mass (something called neutrino oscillation).

The bottom line is, as you've ascertained from that website: theory would be in trouble if the photon had mass. And, this is a well-established theory (quantum electrodynamics), which has stood up to incredible experimental verification, so the chance that something like this could kill it is pretty remote. Unlike with neutrino physics, there is nothing we observe that could be explained by a massive photon. It's more or less a pipe dream of researchers who need papers to write (and want to stir up some trouble).

Again, reporting an upper bound on the mass doesn't mean it has mass. It is just a consequence of experimental verification. We can't measure exactly 0 mass, but the closer we get to 0 the more likely it is 0.



Hope that was good enough.

OK I admit it again. I'm probably even more confused.

The ususal entire equation for E=mc^2
I'm told should instead be Eo = ((Y*eV)/c2) * C^2
So I'm seening "rest energy", which I haven't got to clue what that means, is Equal to some number Y amt times eV/C^2
and all of that is then multiplied together.

The c^2 on the right side of the equation cancel each other out, whcih tells me the equation isn't correct,
and the remaining (Y*eV) is NOW an INVALID, since it sitting over a
one of the 2 C^2, that is now ZERO,, and since you can't divide by zero... IM TOTALLY CONVINCED i DIDN'T UNDERSTAND ANY of the flying equations. What can I say?
I can seem to get a simple equation expressed correctly.
'I still don't know what a zero rest Energy would mean,
if mass is an invariant, why not just call the thing a constant, and
multfy it by the other stuffr you want to multipfy it by.
Yep. REAL LOST. buit willing to listen to anyoen that can put this puuzel together. (Whether any does or not, I be renis if iI did't even thank them for their interest. It seems half of use leaned few new tricks.)

Long story short, if someone can put this thing into a simple equation the way it's suppossed to lock, and also iedentify the varaints) of the equation that would be meighty instructive, nad
even working through 1 example using any number wiul dbe better, cause then I cou;kd ask the the heck Ei <<<< sub 1 really means at as apposed to Eo, and W?HAT is INCREATING, AND WHAT isn't,

 

[selfAdjoint]

When there's a relative speed between you and something else, and both of you are unaccelerated, your physics and their physics are going to have to have conversion factors between themn to be meaningful. Actually, I assume a software developer knows what a matrix is, so what you really truly need to convert is a set of matrices called Lorentz transforms. But in simple cases they can be dummed down to just a factor.

And the most common factor is gamma (you mistook it for a Y), but it's $$\gamma = \frac{1}{\surd(1-\frac{v^2}{c^2})}$$.

Where v is the relative speed and c is the speed of light.

Now there are two systems of applying this factor. In one system we keep energy invariant ("conservation of energy") and apply the factor to mass. In the other we do the opposite, mass is taken as an invariant and energy gets multiplied by gamma. Each system is consistent within itself and unfortunately we have fans of both systems posting on these boards, which really sucks for the newby looking for clarification.

System 1
If the mass you measure in that something is going to vary, then you need to distinuish the mass when it's moving (relative to you) which is multiplied by gamma, from the mass when it's not moving, say relative to itself, when it doesn't have a gamma. So that latter case is called its "rest mass" in the first system I mentioned.

System 2
On the other hand if the mass is invariant, there are not two cases, just one, and mass, or "invariant mass" is always the same. Notice that numerically, invariant mass = rest mass. But as a SW developer you know how little that can mean.

 

[GRQC]

Originally posted by treat2
OK I admit it again. I'm probably even more confused.

The ususal entire equation for E=mc^2
I'm told should instead be Eo = ((Y*eV)/c2) * C^2

No, E=mc². That's it. There's your equation. The units of energy are in eV, and units of mass are in eV/c².

So I'm seening "rest energy", which I haven't got to clue what that means, is Equal to some number Y amt times eV/C^2
and all of that is then multiplied together.

Rest energy 'E' is the "energy equivalent" of a mass 'm'. They're the same thing.

Yep. REAL LOST. buit willing to listen to anyoen that can put this puuzel together. (Whether any does or not, I be renis if iI did't even thank them for their interest. It seems half of use leaned few new tricks.) ...
Long story short, if someone can put this thing into a simple equation the way it's suppossed to lock, and also iedentify the varaints) of the equation that would be meighty instructive, nad
even working through 1 example using any number wiul dbe better, cause then I cou;kd ask the the heck Ei <<<< sub 1 really means at as apposed to Eo, and W?HAT is INCREATING, AND WHAT isn't,

I hope your command of C++ structure is better than your command of English grammar.

If you're not going to take the time to proof-read your post, I'm not going to take the time to answer.

 

[treat2]

Originally posted by GRQC
No, E=mc². That's it. There's your equation. The units of energy are in eV, and units of mass are in eV/c².
Rest energy 'E' is the "energy equivalent" of a mass 'm'. They're the same thing.
...
If you're not going to take the time to proof-read your post, I'm not going to take the time to answer.

Originally posted by selfAdjoint
When there's a relative speed between you and something else, and both of you are unaccelerated, your physics and their physics are going to have to have conversion factors between themn to be meaningful. Actually, I assume a software developer knows what a matrix is, so what you really truly need to convert is a set of matrices called Lorentz transforms. But in simple cases they can be dummed down to just a factor.

And the most common factor is gamma (you mistook it for a Y), but it's $$\gamma = \frac{1}{\surd(1-\frac{v^2}{c^2})}$$.

Where v is the relative speed and c is the speed of light.

Now there are two systems of applying this factor. In one system we keep energy invariant ("conservation of energy") and apply the factor to mass. In the other we do the opposite, mass is taken as an invariant and energy gets multiplied by gamma. Each system is consistent within itself and unfortunately we have fans of both systems posting on these boards, which really sucks for the newby looking for clarification.

System 1
If the mass you measure in that something is going to vary, then you need to distinuish the mass when it's moving (relative to you) which is multiplied by gamma, from the mass when it's not moving, say relative to itself, when it doesn't have a gamma. So that latter case is called its "rest mass" in the first system I mentioned.

System 2
On the other hand if the mass is invariant, there are not two cases, just one, and mass, or "invariant mass" is always the same. Notice that numerically, invariant mass = rest mass. But as a SW developer you know how little that can mean.

Sorry about my previous post. (I can't even read what I wrote. LOL!) Given certain conditions, and a lack of allergy pills, I can become virtually blind, as you can see from my unintelligible post.

I am must misunderstanding the use of the word "invariant", because
some posts here have said the amount of mass never varies, regardless of mass accelerating, or decelerating relative to an observer.

Other posts saying mass is an "invariant", and it varies, according to the mass accelerating or decelerating, relative to an observer outside of that "system".

If the amount of mass varies, in ANY scenario, depending upon acceleration or deceleration, and in what system the observer is in, then:

1) Why refer to mass as an invariant, in ALL systems?
2) Why did your previous posts say that the amount of mass never varies in ANY system?
3) Why refer to something as an invariant, when it does vary?
4) what about your earliest posts in which (I thought) you said that mass is an invariant (i.e. does not vary) relative to an observer in ANY system and in ANY scenario, BUT E does vary?

I'm not trying to be argumentative, but I don't understand why mass is being called an invariant, when it is being said to vary. I am equally confused by what at first appeared to be your position, as I understood it, that mass does NOT vary, relative to anyone, in ANY system, in ANY scenario, and now seems to be saying the opposite?

 

[selfAdjoint]

If something is "invariant" under a system of trasformations, then it will have the same value after you do such a transformation as it did before. For example length is invariant under rotations.

According to ONE of the two ways of formaliziing relativity, mass is invariant and energy is not. According to the OTHER formalism, mass is not invariant but energy is.

If this duality seems counterintuitive think of this. You are looking at something which is speeding by, or toward, or away from you. You observe its behavior, and set out to do equations on that. You find that the equations predict the same behavior if you plug in an invariant mass and an energy that is a certain function of speed, or if you plug in an invariant energy and have the mass be a certain function of the speed. Either way gives you the same predictions of behavior. This is actually a "duality" like the ones the string physicists talk about but SR is such a simple theory the duality has no effect, except to confuse students.

You are going to have to check each answer to your question according to who sent it. Each individual should be consistent with him/herself, but they will disagee with each other, only due to this double formality problem.